在A和B两个地方之间有12个中间车站。请问有多少种方法可以使火车停在这些中间车站中的4个中间车站,这样就不会有两个连续的车站?
例子 –
Input : n = 12, s = 4
Output : 126
Input : n = 16, s = 5
Output : 792
说明1:
固定/删除四个停靠点作为固定点,如果在停靠点之间至少有一个停靠站,则以其他方式计算可以在它们之间插入其他停靠站的方式。
A x x x x x x x x B
在这8个非停止站点之间,我们有9个位置,我们选择这9个位置作为这8个站点之间的停止位置。
因此,答案应该是 = 126
说明2:
如果您知道将难以区别的球组合到不同的盒子中,那么您可以简单地使用, 。在此问题中,$ n $是要停止的站点数,而$ p $是要停止的站点数。在这里,停止站就像无法分辨的球,非停止站就像可区分的盒子。
所以, = = 126
代码 –
C++
#include
int stopping_station(int, int);
// function to calculate number
// of ways of selecting 'p' non consecutive
// stations out of 'n' stations
int stopping_station(int p, int n)
{
int num = 1, dem = 1, s = p;
// selecting 's' positions out of 'n-s+1'
while (p != 1) {
dem *= p;
p--;
}
int t = n - s + 1;
while (t != (n - 2 * s + 1)) {
num *= t;
t--;
}
if ((n - s + 1) >= s)
printf("%d", num / dem);
else
// if conditions does not satisfy of combinatorics
printf("not possible");
}
// driver code
int main()
{
// n is total number of stations
// s is no. of stopping stations
int n, s;
// arguments of function are
// number of stopping station
// and total number of stations
stopping_station(4, 12);
}
Java
// Java code to calculate number
// of ways of selecting 'p' non
// consecutive stations out of
// 'n' stations
import java.io.*;
import java.util.*;
class GFG
{
public static int stopping_station(int p, int n)
{
int num = 1, dem = 1, s = p;
// selecting 's' positions out of 'n-s+1'
while (p != 1)
{
dem *= p;
p--;
}
int t = n - s + 1;
while (t != (n - 2 * s + 1))
{
num *= t;
t--;
}
if ((n - s + 1) >= s)
System.out.print(num / dem);
else
// if conditions does not satisfy of combinatorics
System.out.print("not possible");
return 0;
}
public static void main (String[] args)
{
// n is total number of stations
// s is no. of stopping stations
int n, s;
// arguments of function are
// number of stopping station
// and total number of stations
stopping_station(4, 12);
}
}
// ""This code is contributed by Mohit Gupta_OMG ""
Python3
# Python code to calculate number
# of ways of selecting 'p' non
# consecutive stations out of
# 'n' stations
def stopping_station( p, n):
num = 1
dem = 1
s = p
# selecting 's' positions
# out of 'n-s+1'
while p != 1:
dem *= p
p-=1
t = n - s + 1
while t != (n-2 * s + 1):
num *= t
t-=1
if (n - s + 1) >= s:
return int(num/dem)
else:
# if conditions does not
# satisfy of combinatorics
return -1
# driver code
num = stopping_station(4, 12)
if num != -1:
print(num)
else:
print("Not Possible")
# This code is contributed by "Abhishek Sharma 44"
C#
// C# code to calculate number
// of ways of selecting 'p' non
// consecutive stations out of
// 'n' stations
using System;
class GFG {
public static int stopping_station(int p,
int n)
{
int num = 1, dem = 1, s = p;
// selecting 's' positions
// out of 'n-s+1'
while (p != 1)
{
dem *= p;
p--;
}
int t = n - s + 1;
while (t != (n - 2 * s + 1))
{
num *= t;
t--;
}
if ((n - s + 1) >= s)
Console.WriteLine(num / dem);
// if conditions does not
// satisfy of combinatorics
else
Console.WriteLine("Not possible");
return 0;
}
// Driver Code
public static void Main(String []args)
{
// arguments of function are
// number of stopping station
// and total number of stations
stopping_station(4, 12);
}
}
// This code is contributed by vt_m.
PHP
= $s)
echo $num / $dem;
else
// if conditions does not
// satisfy of combinatorics
echo "not possible";
}
// Driver Code
// n is total number of stations
// s is no. of stopping stations
$n; $s;
// arguments of function are
// number of stopping station
// and total number of stations
stopping_station(4, 12);
// This code is contributed by anuj_67.
?>
Javascript
输出 :
126